2013 AIME I Problems/Problem 15
Contents
Problem 15
Let 
be the number of ordered triples 
of integers satisfying the conditions
(a) 
,
(b) there exist integers 
, 
, and 
, and prime 
where 
,
(c) divides 
, 
, and 
, and
(d) each ordered triple and each ordered triple 
form arithmetic sequences. Find .
Solution 1
From condition (d), we have 
and 
. Condition 
states that 
, 
, and 
. We subtract the first two to get 
, and we do the same for the last two to get 
. We subtract these two to get 
. So 
or 
. The second case is clearly impossible, because that would make 
, violating condition 
. So we have , meaning 
. Condition implies that 
or 
. Now we return to condition , which now implies that 
. Now, we set 
for increasing positive integer values of 
. 
yields no solutions. 
gives 
, giving us 
solution. If 
, we get 
solutions, 
and 
. Proceeding in the manner, we see that if 
, we get 16 solutions. However, 
still gives 
solutions because 
. Likewise, 
gives 
solutions. This continues until 
gives one solution. 
gives no solution. Thus, 
.
Solution 2
Condition c gives us that 
, etc. Condition d then tells us that C and c can be expressed as 
and 
, respectively. However, plugging what we got from condition c into this, we find that 
. From there, we branch off into two cases; either 
, or 
. Realize then that the second case leads to a contradiction, due to condition b. Then, means that must be 
. The bash from there is pretty similar to what was done in Solution 1. We get 
. - Spacesam
See also
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